When the pressure of a gas doubles, the new volume
1 answer:
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Answer:
Volume of = 15.31 mL
Volume of = 4.69 mL
Explanation:
Given that:
the density of the mixture = 1.82 g/mL
From the density of the pure samples
The density of = 1.492 g/mL
The density of = 2.890 g/mL
The total volume of the liquid mixture = 20.0 mL
Suppose the volume of = P ml
and the volume of = Q ml
the sum of their volumes should be equal to the total volume of the mixture
----- (1)
However, we know that Density = mass/volume
∴ mass = density × volume
The equation can now be expressed as:
1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL ---- (2)
From equation (1) ;
let Q = 20 - P
The replace the value of P into equation (2)
1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL
1.492 P g + 57.8g - 2.890 P g = 36.4g
1.492 P g - 2.890 P g = 36.4g - 57.8g
-1.398 P g = -21.4g
P = -21.4g/-1.398g
P = 15.31 mL
Q = 20 - P
Q = (20 - 15.31) mL
Q = 4.69 mL
∴
Volume of = 15.31 mL
Volume of = 4.69 mL
Answer:
Explanation:
Given parameters:
Mass of aluminium oxide = 3.87g
Mass of water = 5.67g
Unknown:
Limiting reactant = ?
Solution:
The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.
Using the number of moles, we can ascertain the limiting reactants;
Al₂O₃ + 3H₂O → 2Al(OH)₃
Number of moles;
Number of moles =
molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole
number of moles = = 0.04mole
molar mass of H₂O = 2(1) + 16 = 18g/mole
number of moles = = 0.32mole
From the reaction equation;
1 mole of Al₂O₃ reacted with 3 moles of H₂O
0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O
But we were given 0.32 mole of H₂O and this is in excess of amount required.
This shows that Al₂O₃ is the limiting reactant